How to solve inequalities $0 \leq \arg\Bigl(\frac {z-i}{z+i}\Bigr) < \frac \pi 4 $ for complex variable $z$

Question

How can I solve the inequalities,

$$ 0 \leq \arg\Bigl(\frac {z-i}{z+i}\Bigr) < \frac \pi 4 $$

for $z$, where $z$ is a complex number.

Answer 1

$0 \le \arg W < \frac {\pi}4$

$0 \le \tan (\arg W) < 1$

$0 \le \frac {Im(W)}{Re(W)} < 1$

Let $Re(z) =a$ and $Im(z) = b$

Then $W= \frac {z+i}{z-i} = \frac {a+i(b+1)}{a+i(b-1)}=$

$\frac {(a+i(b+1))(a-i(b-1))}{(a+i(b-1))(a-i(b-1))} =$

$\frac {[a^2-(b+1)(b-1)] + i[a(b+1)-a(b-1)]}{a^2 + (b-1)^2} =$

$\frac {(a^2 -b^2 + 1) + 2ai}{a^2 + b^2 -2b + 1}$.

So $\frac {Re(W)}{Im(W)} = \frac {2a}{a^2 -b^2 + 1}$

And $0\le \frac {2a}{a^2-b^2 + 1} < 1$

$0 \le 2a < a^2 -b^2 + 1$

$b^2 < a^2 - 2a + 1$

$|b| < |a-1|$

Answer 2

Hint:

$$\arg\frac{z-i}{z+i}=\arg \underbrace{\frac{(z-i)(\bar z-i)}{(z+i)(\bar z-i)}}_{|z+i|^2}=\arg\bigl((z-i)(\bar z-i)\bigr)=\arg\bigl(|z|^2-2i\operatorname{Re}z-1\Bigr).$$ Setting $z=x+iy$, you can expand $$|z|^2-1-2i\operatorname{Re}z=x^2+y^2-1-2ix$$ and translate the inequation as conditions on the tangent and sine of the argument $\theta$: $$0<\tan\theta <1\quad\text{and}\quad \sin\theta >0$$ which are easy to express from $x$and $y$.

Answer 3

Let $z=u+iv$. Calculate

$$\arg\left( \frac{z-i}{z+i} \right) = \arg(z-i)-\arg(z+i) =\tan^{-1}\left( \frac{v-1}{u}\right)-\tan^{-1}\left( \frac{v+1}{u}\right)$$ $$=\tan^{-1}\left( \frac{-2u}{u^2+v^2-1}\right)$$

where the identity $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab}$ is used. Given that $ 0 \leq \arg\Bigl(\frac {z-i}{z+i}\Bigr) < \frac \pi 4 $, we have

$$0\le \frac{-2u}{u^2+v^2-1} <1$$

Solve the inequalities to get the two solutions below

1) $u\ge 0,\>\>\> (u+1)^2+v^2 < 2$;

2) $u\le 0,\>\>\> (u+1)^2+v^2 > 2$.

Visualize the regions of the solutions in the complex plane below.